Animated Solution for Mathematics - Functions: The number of functions $f$, from the set $A=\{x\in \mathbb{N}:x^2-10x+9\le 0\}$ to the set $B=\{n^2:n\in \mathbb{N}\}$ such that $f(x)\le (x-3{)}^2+1$, for every $x\in A$, is

A Journey into Functions

Imagine standing at the threshold of a function machine. You have a set of inputs, $A$, and a set of potential outputs, $B$. Usually, a function is a free-for-all—any input can map to any output.

But today, we are dealing with a constrained system, a puzzle where the rules of the game change depending on where you stand. Let us unravel this together.

Before we can map anything, we must know our domain. We are given $A=\{x\in \mathbb{N}:x^2-10x+9\le 0\}$.

When we factor the quadratic $x^2-10x+9$, we get $(x-1)(x-9)\le 0$. On the number line, this inequality holds between $1$ and $9$.

Since we are restricted to natural numbers, our domain $A$ is simply the set $\{1,2,3,4,5,6,7,8,9\}$. This is our stage.

Our codomain $B$ is the set of all perfect squares: $B=\{1,4,9,16,25,36,\dots \}$.

Now, here is the twist. We are not free to map $x$ to any $y\in B$. We are bound by the condition:

This inequality acts as a gatekeeper. For every $x$ in our domain, it dictates the maximum value that $f(x)$ can take.

We must now make a decision for each $x\in A$. Because each mapping is an independent event, the Fundamental Principle of Counting tells us that the total number of functions is the product of the number of choices for each $x$.

Let us calculate these choices one by one:

For $x=1$: $f(1)\le (1-3{)}^2+1=5$. The perfect squares in $B$ less than or equal to $5$ are $1$ and $4$. That is $2$ choices*.

For $x=2$: $f(2)\le (2-3{)}^2+1=2$. Only $1$ works. That is $1$ choice*.

For $x=3$: $f(3)\le (3-3{)}^2+1=1$. Only $1$ works. That is $1$ choice*.

For $x=4$: $f(4)\le (4-3{)}^2+1=2$. Only $1$ works. That is $1$ choice*.

For $x=5$: $f(5)\le (5-3{)}^2+1=5$. The squares are $1$ and $4$. That is $2$ choices*.

For $x=6$: $f(6)\le (6-3{)}^2+1=10$. The squares are $1,4,9$. That is $3$ choices*.

For $x=7$: $f(7)\le (7-3{)}^2+1=17$. The squares are $1,4,9,16$. That is $4$ choices*.

For $x=8$: $f(8)\le (8-3{)}^2+1=26$. The squares are $1,4,9,16,25$. That is $5$ choices*.

For $x=9$: $f(9)\le (9-3{)}^2+1=37$. The squares are $1,4,9,16,25,36$. That is $6$ choices*.

Now, we simply multiply these independent possibilities:

Calculating this, we get:

There you have it. By breaking down a seemingly complex constraint into individual, manageable decisions, we have navigated the entire problem. The final answer is $1440$.